Arolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m. what is the magnitude of its angular acceleration if there was no slipping?

Refer to the diagram shown below.

r = 68/2 = 34 cm = 0.34 m, the radius of the wheel
u = 8.4 m/s, the intial tangential velocity
s = 115 m, the stopping distance.

The circumference of the wheel is
C = 2πr = 2π(0.34) = 0.17π m
A circumference is equivalent to 2π radians, therefore the stopping distance is equivalent to
θ = [115/(0.17π)]*2π = 215.327 rad

The intial angular velocity is
ω₀ = u/r = (8.4 m/s)/(0.34 m) = 24.706 rad/s

Let α =  the angular acceleration.
Because the wheel comes to rest, therefore
ω₀² + 2*α*θ = 0
(24.706 rad/s)² +2*(α rad/s)*(215.327 rad) = 0
610.3815 + 430.654α = 0
α = - 1.417 rad/s²

-1.417 rad/s²

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